I wanted to ask a question inspired by an excellent answer to the query about the intuition for the beta distribution. I wanted to get a better understanding of the derivation for the prior distribution for the batting average. It looks like David is backing out the parameters from the mean and the range.

Under the assumption that the mean is $0.27$ and the standard deviation is $0.18$, can you back out $\alpha$ and $\beta$ by solving these two equations: \begin{equation} \frac{\alpha}{\alpha+\beta}=0.27 \\ \frac{\alpha\cdot\beta}{(\alpha+\beta)^2\cdot(\alpha+\beta+1)}=0.18^2 \end{equation}

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    $\begingroup$ Honestly, I just kept graphing values in R until it looked right. $\endgroup$ Commented Jan 17, 2013 at 3:09
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    $\begingroup$ where do you get the standard deviation to be .18? $\endgroup$
    – appleLover
    Commented Jul 20, 2013 at 19:20
  • $\begingroup$ How did you come up with this standard deviation? Did you know it in advance? $\endgroup$
    – Daria
    Commented Jun 12, 2019 at 6:02

2 Answers 2


Notice that:

\begin{equation} \frac{\alpha\cdot\beta}{(\alpha+\beta)^2}=(\frac{\alpha}{\alpha+\beta})\cdot(1-\frac{\alpha}{\alpha+\beta}) \end{equation}

This means the variance can therefore be expressed in terms of the mean as

\begin{equation} \sigma^2=\frac{\mu\cdot(1-\mu)}{\alpha+\beta+1} \\ \end{equation}

If you want a mean of $.27$ and a standard deviation of $.18$ (variance $.0324$), just calculate:

\begin{equation} \alpha+\beta=\frac{\mu(1-\mu)}{\sigma^2}-1=\frac{.27\cdot(1-.27)}{.0324}-1=5.083333 \\ \end{equation}

Now that you know the total, $\alpha$ and $\beta$ are easy:

\begin{equation} \alpha=\mu(\alpha+\beta)=.27 \cdot 5.083333=1.372499 \\ \beta=(1-\mu)(\alpha+\beta)=(1-.27) \cdot 5.083333=3.710831 \end{equation}

You can check this answer in R:

> mean(rbeta(10000000, 1.372499, 3.710831))
[1] 0.2700334
> var(rbeta(10000000, 1.372499, 3.710831))
[1] 0.03241907
  • $\begingroup$ David, do you happen to follow any baseball research? There are several competing techniques out there for finding the proper $\alpha$ and $\beta$, so I was wondering if you had any opinion on the matter if you were doing something besides just trying to find a graph that looked reasonable. $\endgroup$ Commented Jan 17, 2013 at 14:53
  • $\begingroup$ I don't particularly follow sabermetrics- in the other answer it just happened to provide a very convenient example of estimating p from a binomial with a prior. I don't even know if this is how it's done in sabermetrics, and if it is, I know there are many components I left out (players having different priors, stadium adjustments, weighting recent hits over old ones...) $\endgroup$ Commented Jan 17, 2013 at 15:02
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    $\begingroup$ I am impressed that your eyeballing was this accurate. $\endgroup$
    – dimitriy
    Commented Jan 19, 2013 at 1:03
  • $\begingroup$ Hi David, how do you get from these values of $\alpha = 1.37$ and $\beta = 3.71$ to your eyeballed values in the linked post of 81 and 219 respectively? $\endgroup$
    – Alex
    Commented Feb 29, 2016 at 2:15
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    $\begingroup$ @Alex The requested variance and standard deviation come from the above question, which asked for an SD of .18, not the beta distribution post. If I were calculating instead of eyeballing I might have guessed an SD of something like .03, which would have given values of 59 and 160. $\endgroup$ Commented Feb 29, 2016 at 2:32

I wanted to add this as a comment on the excellent answer but it ran long and will look better with answer formatting.

Something to keep in mind is that not all $(\mu, \sigma^2)$ are possible. It's clear $\mu \in [0,1]$, but not as clear are the limitations for $\sigma^2$.

Using the same reasoning as David, we can express

$$ \sigma^2(\alpha, \mu) = \frac{\mu^2 (1-\mu)}{\alpha + \mu} $$

This is decreasing with respect to $\alpha$, so the largest $\sigma^2$ can be for a given $\mu$ is:

$$\lim_{\alpha \rightarrow 0}\sigma^2(\alpha, \mu) = \mu(1-\mu)$$

This is only a supremum since the set of valid $\alpha$ is open (i.e., for Beta, we must have $\alpha > 0$); this limit is itself maximized at $\mu = \frac12$.

Notice the relationship to a corresponding Bernoulli RV. The Beta distribution with mean $\mu$, since it is forced to take all values between 0 and 1, must be less dispersed (i.e., have lower variance) than the Bernoulli RV with the same mean (which has all of its mass at the ends of the interval). In fact, sending $\alpha$ to 0 and fixing $\beta = \frac{1-\mu}{\mu} \alpha$ amounts to putting more and more of the mass of the PDF close to 0 and 1, i.e., getting closer to a Bernoulli distribution, which is why the supremum of the variance is exactly the corresponding Bernoulli variance.

Taken together, here is the set of valid means and variances for Beta:

enter image description here

(Indeed this is noted on the Wikipedia page for Beta)


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